600 3. DETERMINING THE CASES FOR LE .Cj(G, T)
hence A(T) = {A, V} is of order 2. Therefore as Dis of odd order, D acts on V,
contrary to 3.3.6.b. So (1) is estabished.
Finally we prove (3), so we assume that CT(L) = 1, and V < 02 (LT). By (1),
we may assume Lis L 3 (2) or U 3 (3), and it remains to derive a contradiction. As
CT(L) = 1, Q := 02 (LT) is elementary abelian by C.1.13.a. Further by C.1.13.b,
B.4.8, and B.4.6, Q is the indecomposable module with natural irreducible sub-
module V and trivial quotient, of rank 4 or 7, respectively. By 3.3.7.2, V is an
FF-module, so by B.4.6.13, Lis not U 3 (3). Thus L ~ L3(2), and by B.4.8.3, there
is a unique member A of P(T, Q). As CQ(A) = CQ(a) for each a E _A# and
Q = CLT(Q), we may apply B.2.21 to obtain the same contradiction as earlier.
This completes the proof of (3). D
Observe now that as L is a block by 3.3.25.1, Hypothesis C.6.2 is satisfied with
L, T, T, TD in the roles of "L, R, TH, A", For example, if 1 =!= Ro ~ T with
Ro ~ LT, then D i. Nc(Ro) by 3.3.6.b, which verifies part (3) of Hypothesis
C.6.2. As Hypothesis C.6.2 is satisfied, we can apply C.6.3 to conclude:LEMMA 3.3.26. There exists d ED - M with Vd i. 02(LT).
In the remainder of the section, let d be defined as in 3.3.26. Set QL := 02(LT)
and Tc:= CT(L).
LEMMA 3.3.27. Assume Tc= CT(L) =/= 1. Then
{1) Tc n rt;= 1.
{2) <I>(Tc) = 1.(3) Either Tf; ~ QL or Tc~ Q1.
PROOF. As L centralizes Tc ~ T and D acts on T, also Tf; ~ T, and then
Tc n Tf; is normal in LT and in LdT. Thus if Tc n Tf; =/= 1, then
Md= !M(LdT) = !M(Nc(Tc n rt;)) = !M(LT) = M,
contradicting our choice of d E D - M in 3.3.26, and so establishing (1). Then
applying (1) to d^2 in the role of "d", Tc n rtJ = 1, so also Tf;-
1
n Tf; = 1.Now as L is a block, (QL) ~ Tc by C.1.13.a. Suppose (2) fails, so that
<!>(Tc) =/= 1. If Tf; ~ QL, then 1 =/= <.P(Tf;) ~ <.P(QL) ~ Tc, contradicting (1);
therefore Tf; i. QL, so by symmetry Tc i. Q1, and thus (3) fails. Hence (3) implies(2), so it remains to assume that (3) fails, and to derive a contradiction. Thus
Tf; i. QL and Tc i. Q1, so also Tf;-
1
i. QL.
Suppose for the moment that L is L 3 (2). Then by 3.3.23.1, D acts on the
preimage To in T of Z(T). Therefore as '1' 0 is of order 2 and Tf; i. QL, 'f'tJ = T 0.Now suppose that Lis not L3(2). Then by 3.3.25.2, D acts on X := 02 (CL(Z))
and on R := 02(XT). Therefore as Tc ~ XT, T!; ~ XT, and as Tc centralizes
X, 1 =/= 'f'!J centralizes X. Now L ~ A 6 or G 2 (2)' by 3.3.25.1, and Tis trivial on
the Dynkin diagram of L if L ~ A5 since Lis an A5-block. Inspecting Aut(L), we
find that CAut(L)(X) = 1 unless LT~ 85, whereas we saw 'ff;=/= 1 centralizes X.
Therefore LT~ 86 and 'ff;= Z(XT) = T 0 is of order 2.
Thus LT~ 85 or L3(2) and 'ff;= T 0 is of order 2. As Tf; ~ T, 1 =/= [V, T 0 ] =
[ d] d.. [ d-(^1) d- (^1) d d- 1
V, Tc ~ Tc· S1m1larly V, To] = [V, Tc ] ~ Tc , so 1 =I= [V, T 0 ] ~ Tc n Tc ,
contrary to the final remark in paragraph one. D